What is the tangent of f(x)=-x^2+4sqrt(x) ?

The tangent of f(x)=-x^2+4sqrt(x) is y=(-2a_{0}+2/(sqrt(a_{0))})x+a_{0}^2+2sqrt(a_{0)}

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tangent of f(x)=-x^2+4sqrt(x)

tangent of

f (x) = - x^{2} + 4 x

y = (- 2 a_{0} + \frac{2}{a _{0}}) x + a_{0}^{2} + 2 a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - a_{0}^{2} + 4 a_{0})

Find the slope of

f (x) = - x^{2} + 4 x : \frac{df ( x )}{d x} = - 2 x + \frac{2}{x}

Compute the slope of

f (x) = - x^{2} + 4 x

x = a_{0} : m = - 2 a_{0} + \frac{2}{a _{0}}

Find the line with slope m=

- 2 a_{0} + \frac{2}{a _{0}}

and passing through

(a_{0}, - a_{0}^{2} + 4 a_{0}) : y = (- 2 a_{0} + \frac{2}{a _{0}}) x + a_{0}^{2} + 2 a_{0}

y = (- 2 a_{0} + \frac{2}{a _{0}}) x + a_{0}^{2} + 2 a_{0}

\int \frac{10 x - 2}{5 x ^{2} - 2 x} d x

\int \frac{( 1 + sin ( x ) )}{cos ^{2} ( x )} d x

x \to 0 - lim (csc (x))

y^{^{''}} - 25 y = 0, y (2) = - 2, y^{^{'}} (- 2) = 2

y^{^{'}} = y cos (t)

What is the tangent of f(x)=-x^2+4sqrt(x) ?
The tangent of f(x)=-x^2+4sqrt(x) is y=(-2a_{0}+2/(sqrt(a_{0))})x+a_{0}^2+2sqrt(a_{0)}